3.940 \(\int \frac{(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=114 \[ -\frac{6 i a^4}{f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac{i a^4 \log (\cos (e+f x))}{c^3 f}-\frac{a^4 x}{c^3}+\frac{6 i a^4}{c f (c-i c \tan (e+f x))^2}-\frac{8 i a^4}{3 f (c-i c \tan (e+f x))^3} \]

[Out]

-((a^4*x)/c^3) + (I*a^4*Log[Cos[e + f*x]])/(c^3*f) - (((8*I)/3)*a^4)/(f*(c - I*c*Tan[e + f*x])^3) + ((6*I)*a^4
)/(c*f*(c - I*c*Tan[e + f*x])^2) - ((6*I)*a^4)/(f*(c^3 - I*c^3*Tan[e + f*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.136389, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {3522, 3487, 43} \[ -\frac{6 i a^4}{f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac{i a^4 \log (\cos (e+f x))}{c^3 f}-\frac{a^4 x}{c^3}+\frac{6 i a^4}{c f (c-i c \tan (e+f x))^2}-\frac{8 i a^4}{3 f (c-i c \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^4/(c - I*c*Tan[e + f*x])^3,x]

[Out]

-((a^4*x)/c^3) + (I*a^4*Log[Cos[e + f*x]])/(c^3*f) - (((8*I)/3)*a^4)/(f*(c - I*c*Tan[e + f*x])^3) + ((6*I)*a^4
)/(c*f*(c - I*c*Tan[e + f*x])^2) - ((6*I)*a^4)/(f*(c^3 - I*c^3*Tan[e + f*x]))

Rule 3522

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx &=\left (a^4 c^4\right ) \int \frac{\sec ^8(e+f x)}{(c-i c \tan (e+f x))^7} \, dx\\ &=\frac{\left (i a^4\right ) \operatorname{Subst}\left (\int \frac{(c-x)^3}{(c+x)^4} \, dx,x,-i c \tan (e+f x)\right )}{c^3 f}\\ &=\frac{\left (i a^4\right ) \operatorname{Subst}\left (\int \left (\frac{1}{-c-x}+\frac{8 c^3}{(c+x)^4}-\frac{12 c^2}{(c+x)^3}+\frac{6 c}{(c+x)^2}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^3 f}\\ &=-\frac{a^4 x}{c^3}+\frac{i a^4 \log (\cos (e+f x))}{c^3 f}-\frac{8 i a^4}{3 f (c-i c \tan (e+f x))^3}+\frac{6 i a^4}{c f (c-i c \tan (e+f x))^2}-\frac{6 i a^4}{f \left (c^3-i c^3 \tan (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 2.37909, size = 143, normalized size = 1.25 \[ \frac{a^4 (\cos (3 e+7 f x)+i \sin (3 e+7 f x)) \left (-9 \sin (e+f x)+6 i f x \sin (3 (e+f x))+2 \sin (3 (e+f x))-3 i \cos (e+f x)+\cos (3 (e+f x)) \left (3 i \log \left (\cos ^2(e+f x)\right )-6 f x-2 i\right )+3 \sin (3 (e+f x)) \log \left (\cos ^2(e+f x)\right )\right )}{6 c^3 f (\cos (f x)+i \sin (f x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^4/(c - I*c*Tan[e + f*x])^3,x]

[Out]

(a^4*((-3*I)*Cos[e + f*x] + Cos[3*(e + f*x)]*(-2*I - 6*f*x + (3*I)*Log[Cos[e + f*x]^2]) - 9*Sin[e + f*x] + 2*S
in[3*(e + f*x)] + (6*I)*f*x*Sin[3*(e + f*x)] + 3*Log[Cos[e + f*x]^2]*Sin[3*(e + f*x)])*(Cos[3*e + 7*f*x] + I*S
in[3*e + 7*f*x]))/(6*c^3*f*(Cos[f*x] + I*Sin[f*x])^4)

________________________________________________________________________________________

Maple [A]  time = 0.031, size = 91, normalized size = 0.8 \begin{align*} 6\,{\frac{{a}^{4}}{f{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) }}-{\frac{6\,i{a}^{4}}{f{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}-{\frac{i{a}^{4}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{f{c}^{3}}}-{\frac{8\,{a}^{4}}{3\,f{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^4/(c-I*c*tan(f*x+e))^3,x)

[Out]

6/f*a^4/c^3/(tan(f*x+e)+I)-6*I/f*a^4/c^3/(tan(f*x+e)+I)^2-I/f*a^4/c^3*ln(tan(f*x+e)+I)-8/3/f*a^4/c^3/(tan(f*x+
e)+I)^3

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^4/(c-I*c*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [A]  time = 1.30264, size = 193, normalized size = 1.69 \begin{align*} \frac{-2 i \, a^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + 3 i \, a^{4} e^{\left (4 i \, f x + 4 i \, e\right )} - 6 i \, a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i \, a^{4} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{6 \, c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^4/(c-I*c*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

1/6*(-2*I*a^4*e^(6*I*f*x + 6*I*e) + 3*I*a^4*e^(4*I*f*x + 4*I*e) - 6*I*a^4*e^(2*I*f*x + 2*I*e) + 6*I*a^4*log(e^
(2*I*f*x + 2*I*e) + 1))/(c^3*f)

________________________________________________________________________________________

Sympy [A]  time = 1.86559, size = 138, normalized size = 1.21 \begin{align*} \frac{i a^{4} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{3} f} + \frac{\begin{cases} - \frac{i a^{4} e^{6 i e} e^{6 i f x}}{3 f} + \frac{i a^{4} e^{4 i e} e^{4 i f x}}{2 f} - \frac{i a^{4} e^{2 i e} e^{2 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (2 a^{4} e^{6 i e} - 2 a^{4} e^{4 i e} + 2 a^{4} e^{2 i e}\right ) & \text{otherwise} \end{cases}}{c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**4/(c-I*c*tan(f*x+e))**3,x)

[Out]

I*a**4*log(exp(2*I*f*x) + exp(-2*I*e))/(c**3*f) + Piecewise((-I*a**4*exp(6*I*e)*exp(6*I*f*x)/(3*f) + I*a**4*ex
p(4*I*e)*exp(4*I*f*x)/(2*f) - I*a**4*exp(2*I*e)*exp(2*I*f*x)/f, Ne(f, 0)), (x*(2*a**4*exp(6*I*e) - 2*a**4*exp(
4*I*e) + 2*a**4*exp(2*I*e)), True))/c**3

________________________________________________________________________________________

Giac [A]  time = 1.5911, size = 263, normalized size = 2.31 \begin{align*} -\frac{\frac{60 i \, a^{4} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{c^{3}} - \frac{30 i \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c^{3}} - \frac{30 i \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c^{3}} + \frac{-147 i \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 1002 \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 2445 i \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 3820 \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2445 i \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1002 \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 147 i \, a^{4}}{c^{3}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{6}}}{30 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^4/(c-I*c*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-1/30*(60*I*a^4*log(tan(1/2*f*x + 1/2*e) + I)/c^3 - 30*I*a^4*log(abs(tan(1/2*f*x + 1/2*e) + 1))/c^3 - 30*I*a^4
*log(abs(tan(1/2*f*x + 1/2*e) - 1))/c^3 + (-147*I*a^4*tan(1/2*f*x + 1/2*e)^6 + 1002*a^4*tan(1/2*f*x + 1/2*e)^5
+ 2445*I*a^4*tan(1/2*f*x + 1/2*e)^4 - 3820*a^4*tan(1/2*f*x + 1/2*e)^3 - 2445*I*a^4*tan(1/2*f*x + 1/2*e)^2 + 1
002*a^4*tan(1/2*f*x + 1/2*e) + 147*I*a^4)/(c^3*(tan(1/2*f*x + 1/2*e) + I)^6))/f